Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x+8y &= 3 \\ -4x-4y &= -1\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-4y = 4x-1$ Divide both sides by $-4$ to isolate $y$ $y = {-x + \dfrac{1}{4}}$ Substitute this expression for $y$ in the first equation. $5x+8({-x + \dfrac{1}{4}}) = 3$ $5x - 8x + 2 = 3$ Simplify by combining terms, then solve for $x$ $-3x + 2 = 3$ $-3x = 1$ $x = -\dfrac{1}{3}$ Substitute $-\dfrac{1}{3}$ for $x$ back into the top equation. $5( -\dfrac{1}{3})+8y = 3$ $-\dfrac{5}{3}+8y = 3$ $8y = \dfrac{14}{3}$ $y = \dfrac{7}{12}$ The solution is $\enspace x = -\dfrac{1}{3}, \enspace y = \dfrac{7}{12}$.